Question

Calculate the molarity of the solution produced by mixing 31.0 mL of 0.180 M NaCl and 80.0 mL of 0.150 M NaCl. (Assume that the volumes are additive.)

Answer 1

`M=0.1584M`

Step-by-step explanation:

Hello,

In this case, a mixing process is carried out, so the molarity should be modified by considering the new volume of the solution after mingling, in such a way, the resulting moles are:

`n_(NaCl)=0.031L*0.180molNaCl/L+0.080L*0.150molNaCl/L\\n_(NaCl)=0.01758molNaCl`

Afterwards, the total new volume of the solution is:

`V_f=31.0mL+80.0mL=111.0mL*(1L)/(1000mL)=0.111L`

Finally, the new molality is:

`M=(0.01758mol)/(0.111L)=0.1584M`

Best regards.